The least number which when divided by 2,3, 4,5 and 6 leaves the remainder 1 in each case. If the same number is divided by 7 it leaves no remainder. The number is :

Options

A. 231
B. 301
C. 371
D. 441

Answer

To find the least number which, when divided by 2, 3, 4, 5, and 6 leaves a remainder of 1, and when divided by 7 leaves no remainder, we can follow these steps:

### Step 1: Find the LCM of 2, 3, 4, 5, and 6
To find the least common multiple (LCM), we need to consider the highest powers of each prime factor present in the numbers:

- **2**: \(2^2\) (from 4)
- **3**: \(3^1\)
- **5**: \(5^1\)

Thus, the LCM is calculated as:
\[ \text{LCM} = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \]

### Step 2: Formulate the number
Since the number leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6, we can express the number as:
\[ N = 60k + 1 \]
where \(k\) is a non-negative integer.

### Step 3: Ensure \(N\) is divisible by 7
We need \(N\) to be divisible by 7:
\[ 60k + 1 \equiv 0 \ (\text{mod} \ 7) \]
This simplifies to:
\[ 60k \equiv -1 \ (\text{mod} \ 7) \]
Calculating \(60 \mod 7\):
\[ 60 \div 7 = 8 \quad \text{(remainder 4)} \]
So, \(60 \equiv 4 \ (\text{mod} \ 7)\). Thus, we rewrite the equation:
\[ 4k \equiv -1 \ (\text{mod} \ 7) \]
This can be rewritten as:
\[ 4k \equiv 6 \ (\text{mod} \ 7) \]

### Step 4: Solve for \(k\)
To solve \(4k \equiv 6 \ (\text{mod} \ 7)\), we can test values for \(k\):

- For \(k = 0\): \(4(0) \equiv 0\)
- For \(k = 1\): \(4(1) \equiv 4\)
- For \(k = 2\): \(4(2) \equiv 1\)
- For \(k = 3\): \(4(3) \equiv 5\)
- For \(k = 4\): \(4(4) \equiv 2\)
- For \(k = 5\): \(4(5) \equiv 6\) (this works)

Thus, \(k = 5\) is a solution.

### Step 5: Calculate \(N\)
Now substituting \(k = 5\) back into the equation for \(N\):
\[ N = 60(5) + 1 = 300 + 1 = 301 \]

### Conclusion
The least number that satisfies the conditions is:
\[ \boxed{301} \]

---